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设数列{An}(n=1,2,…)是等差数列,且公差为D,...

(1)证明:an=4+(n-1)?2=2n+2,对任意的m,n∈N*,有am+an=(2m+2)+(2n+2)=2(m+n+1)+2,∵m+n+1∈N+,于是,令p=m+n+1,则有ap=2p+2∈{an}.∴该数列是“封闭数列”.(2)∵a1=-5,a2=-3,∴a1+a2=-8,令an=a1+a2=-8,∴2n-7=-8,n=-12?N+,所以...

解: am=Sm-S(m-1)=0-(-9)=9 Sm=(a1+am)m/2=0 a1+am=0 a1=-am=-9 又am=a1+(m-1)d d=(am-a1)/(m-1)=[9-(-9)]/(m-1)=18/(m-1) m>3,18/(m-1)

(1)①当n=4时,a1,a2,a3,a4中不可能删去首项或末项,否则等差数列中连续三项成等比数列,则推出d=0.若删去a2,则a32=a1?a4,即(a1+2d)2=a1?(a1+3d)化简得a1+4d=0,得a1d=?4若删去a3,则a22=a1?a4,即(a1+d)2=a1?(a1+3d)化简得a1-d...

由数列{an}是公差为d,且首项为a0=d的等差数列得:an=a0+(n+1-1)d=(n+1)d;∴Sn+1=a0C0n+a1C1n+…+anCnn,又Sn+1=anCnn+an?1Cn?1n+…+a0C0n=anC0n+an?1C1n+…+a0Cnn,∴2Sn+1=(a0+an)C 0n+(a1+an?1)C1n+…+(an+a0)Cnn=(a0+an)(C0n+C1n+…+Cnn)=...

(1)∵an+1=2Sn+1,∴当n≥2时,an=2Sn-1+1两式相减得:an+1=3an(n≥2)又a2=2a1+1=3=3a1,∴an+1=3an(n∈N*).∴数列{an}是以1为首项,3为公比的等比数列,∴an=3n-1.又b1=b2-d=5-2=3,∴bn=b1+(n-1)d=2n-1.(2)an?bn=(2n+1)?3n?1令Tn=3×1+5×...

(1)由an+1=2Sn+3,得an=2Sn-1+3(n≥2)…(2分)相减得:an+1-an=2(Sn-Sn-1),即an+1=3an,∵当n=1时,a2=2a1+3=9,∴a2a1=3,∴数列{an}是等比数列,∴an=3?3n-1=3n…(5分)(2)∵b1+b2+b3=15,b1+b3=2b2,∴b2=5…(6分)由题意,a13+b1,a23+b2...

(1)①由题意,得2a1+4d=178a1+28d=56,解得d=-1…(4分)②由①知a1=212,所以an=232?n,则bn=3n?an=3n?(232?n),…(6分)因为bn+1-bn=2×3n×(10-n)…(8分)所以b11=b10,且当n≤10时,数列{bn}单调递增,当n≥11时,数列{bn}单调递减,故当n...

(1)∵a10=5,d=2,∴an=2n-15.又∵b3=4,q=2,∴bn=2n-1,∴cn=(2n-15)?2n-1.(2)∵Sn=c1+c2+c3+…+cn,∴2Sn=2c1+2c2+2c3+…+2cn,错位相减,得-Sn=c1+(c2-2c1)+(c3-2c2)+…+(cn-2cn-1)-2cn.∵c1=-13,cn-2cn-1=2n,∴-Sn=-13+22+23+…+2n-(2...

(Ⅰ)∵a2是a1与a4的等比中项,∴a22=a1a4,∵在等差数列{an}中,公差d=2,∴(a1+d)2=a1(a1+3d),即(a1+2)2=a1(a1+3×2),化为2a1=22,解得a1=2.∴an=a1+(n-1)d=2+(n-1)×2=2n.(Ⅱ)∵bn=a n(n+1)2=n(n+1),∴Tn=-b1+b2-b3+b4-…+(-1)nbn=-1...

证明的技巧性较高,用到了均值不等式和部分放缩法,有些难度。

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